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Source: Ireland 2001

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#1 h Jul 5, 2009, 7:31 PM • 3 Y

Y by mathematicsy, Adventure10, Mango247

In an acute-angled triangle $ABC$ , $D$ is the foot of the altitude from $A$ , and $P$ a point on segment $AD$ . The lines $BP$ and $CP$ meet $AC$ and $AB$ at $E$ and $F$ respectively. Prove that $AD$ bisects the angle $EDF$ .

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#2 h Jul 5, 2009, 8:29 PM • 2 Y

Y by Adventure10, Mango247

This is a particular case of the following fact:

$P$ is a point on the plane of $\triangle ABC.$ Lines $PA,PB,PC$ cut $BC,CA,AB$ at $D,E,F.$ Define $U \equiv EF \cap AD$ and let $V$ be the orthogonal projection of $U$ onto $BC.$ Then lines $VU,BC$ bisect $\angle FVE.$

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#3 h Jul 6, 2009, 2:01 AM • 2 Y

Y by Adventure10, Mango247

Let $H_a\equiv EF\cap BC$ . It is well- known that $(H_aDBC)=-1$ . Let $G_a\equiv EF\cap AD$ $\Longrightarrow (H_aG_aEF)=-1$ . Hence $D(H_aG_aEF)=-1$ , but $DH_a\perp DG_a$ , thus $DG_a$ is the internal bisector of $\angle EDF$ $\square$

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#4 h Aug 4, 2010, 5:31 PM • 2 Y

Y by Adventure10, Mango247

Let $DF$ and $DE$ intersect the parallel through $A$ to $BC$ at $X$ and $Y$ respectively and, as $XY\parallel BC\perp AD$ , we need to prove $AX=AY$ .
But $\frac{AX}{BD}=\frac{AF}{BF}$ and $\frac{AY}{DC}=\frac{AE}{CE}$ , or, dividing side by side the two equalities: $\frac{AX}{AY}=\frac{AF\cdot BD}{BF}\cdot\frac{CE}{AE\cdot DC}$ , but, as from Ceva for $\triangle ABC$ with $P$ , the right side of the last equality is $1$ , hence our goal has been achieved.

Best regards,
sunken rock

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#5 h Jul 12, 2020, 3:35 PM • 1 Y

Y by Kgxtixigct

Solution with Complex Bash:

Let $D = 0$ , $A = ai$ , $B = b$ , $C = c$ and $P = pi$ .

$E$ lies on $AC$ and $BP$ , so for some $\lambda, \gamma \in \mathbb{R}$ ,
$E = c + \lambda(ai - c) = b + \gamma(pi - b)$ Comparing real and imaginary coefficients, we have $\gamma = \frac{\lambda a}{p}$ and $\lambda = \frac{cp - pb}{cp - ab}$ , so
$E = c + \left(\frac{cp - pb}{cp - ab}\right) \cdot (ai - c).$ Similarly,
$F = b + \left(\frac{bp - pc}{bp - ac}\right)\cdot (ai - b).$
We wish to show $\angle EDA = \angle ADF$ , that is (with directed angles)
$\begin{align*} \operatorname{Arg}\left(\frac{D - E}{D - A}\right) &= \operatorname{Arg}\left(\frac{D - A}{D - F}\right)\\ \iff \operatorname{Arg}\left(\frac{E}{ai}\right) &= \operatorname{Arg}\left(\frac{ai}{F}\right)\\ \iff \operatorname{Arg}\left(E\right) - \frac{\pi}{2} &= -\operatorname{Arg}\left(F\right)+ \frac{\pi}{2}\\ \iff \operatorname{Arg}\left(E\cdot F\right) &=\pi, \end{align*}$ so it suffices to show that $E\cdot F$ is real, and expanding,
$\begin{align*} E\cdot F &= \left(c + \left(\frac{cp - pb}{cp - ab}\right) \cdot (ai - c)\right) \cdot \left( b + \left(\frac{bp - pc}{bp - ac}\right)\cdot (ai - b)\right) \\ &= \frac{(-a b c-i a b p+i a c p+b c p)(-a b c+i a b p-i a c p+b c p)}{(c p-a b)(b p-a c)}\\ &=\frac{a^{2} b^{2} c^{2}+a^{2} b^{2} p^{2}-2 a^{2} b c p^{2}+a^{2} c^{2} p^{2}-2 a b^{2} c^{2} p+b^{2} c^{2} p^{2}}{a^{2} b c-a b^{2} p-a c^{2} p+b c p^{2}}, \end{align*}$ which is real, so this holds.

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#6 h Jul 12, 2020, 3:36 PM

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(Am I correct in saying that we can say $\operatorname{Arg}(E\cdot F) = \pi$ is equivalent to saying $E \cdot F$ is real as we use directed angles and won't have configuration issues because of the acute $ABC$ ? If not, you'd probably have to prove that it's negative too)

This post has been edited 1 time. Last edited by circlethm, Jul 12, 2020, 3:37 PM
Reason: added more detail

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